Find the xvalue(s) of the relative maxima and relative minima
Calculus

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First, we rewrite the 32/x by using this power law: 1/x = 1/x^1 = x^1
f(x) = 2x^2 + 32/x + 7 > f(x) = 2x^2 + 32x^1 + 7
Then we find the derivative of the function. Let's remember the derivative of a power: (x^n)' = n*x^n1 ; where n is the exponent of the x. By the way, let's remember also that the derivative of any constant is 0.
f'(x) = (2x^2 + 32x^1 + 7)' > f'(x) = (2x^2)' + (32x^1)' + (7)' > f'(x) = 2(x^2)' + 32(x^1)' + (7)'
f'(x) = 2(2x^21) + 32(1x^11) + 0 > f'(x) = 4x^1  32x^2 > f'(x) = 4x  32x^2
f'(x)= 4x  32/x^2
Then we equal the derivative to zero and solve for x.
4x  32/x^2 = 0 > 4x*x^2  32/x^2 * x^2 = 0 > 4x^3  32 = 0 > 4x^3  32 + 32 = 0 + 32
4x^3 = 32 > 4x^3/4 = 32/4 > x^3 = 8 > (x^3)^1/3 = (8)^1/3 > x = 8^1/3
Or x = cubic root(8) > x = 2
Then we find the second derivative.
f''(x) = (4x  32x^2)' = (4x)'  (32x^2)' = 4(x)'  32(x^2)' = 4(1)  32(2)x^21 = 4 + 64x^3
f''(x) = 4 + 64/x^3
Then we enter x = 2 into the second derivative and if it is greater than zero (positive) then it is a minimum
and if it is less than zero (negative) then it is a maximum.
f''(2) = 4 + 64/2^3 = 4 + 64/8 = 4 + 8 = 12 > 0 (positive). Then x = 2 is a minimum and we don't have any maximum.
Relative maxima: DNE (we don't have it).
Relative minima: x = 2
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